Show that the symmetric group $S_p=\left<\sigma, \tau\right>$, where $\sigma$ is any transposition and $\tau$ is any p-cycle and p is a prime number.

Question

Let $\sigma = (a_1\ a_2) \ \ and \ \ \tau = (a_1\ b_2\ \ldots\ b_p)$. (We have $a_2 = b_i$ for some $i$.) We know that $S_p$ is generated by $(a_1\ a_2)$ and $(a_1\ a_2\ \ldots\ a_p)$. I want to show that $\tau^k(a_1) = a_2$ for some $k$. Please help me.

Thank you

I'm not sure I understand your question. If $\tau = (a_1\ b_2\ \dots b_p)$ then $\tau^{j}(a_1) = b_{j+1}$; if $b_i = a_2$, then $\tau^{i-1}(a_1) = b_i = a_2$. Is that what you're after? — Michael Albanese, Jul 25 '14 at 12:16
@ Michael Albanese : I am also thinking as like you, but this not valid for every integer. — Struggler, Jul 25 '14 at 12:20
But $4$ is not prime. The question you asked has been correctly answered, so what is the problem? — Derek Holt, Jul 25 '14 at 12:25
Derke Holt : I am saying that This is valid for a prime number. — Struggler, Jul 25 '14 at 12:31
@johannesvalks: If $\tau = (a_1,b_2,\ldots,b_p)$, then $\tau^j(a_1) = b_{j+1}$ for $1 \le j < p$. That statement is true for any $p$, not just for $p$ prime. That was the question that was asked. — Derek Holt, Jul 25 '14 at 12:39
@ Derk Holt : How to show that $\tau ^k(a_1) = a_2$ for some k. — Struggler, Jul 25 '14 at 12:42
That was answered in the original comment of Michael Albanese. And that statement is true for all $p$, not just for $p$ prime. — Derek Holt, Jul 25 '14 at 12:43
I wonder if the statement is general true...
$< \sigma, \tau >$ is a dihedral group, and only $D_3 = S_3$. — johannesvalks, Jul 25 '14 at 12:43
@ohannesvalks $\langle \sigma,\tau \rangle$ is not a dihedral group in general: only when $p=3$. — Derek Holt, Jul 25 '14 at 12:50
@DerekHolt: I think that $<\tau,\sigma>$ does not generate $S_5$. — johannesvalks, Jul 25 '14 at 12:50
@johannesvalks: It is a standard result, and not hard to prove, that $\langle (1,2),(1,2,3,\ldots,n) \rangle = S_n$ for all $n$. — Derek Holt, Jul 25 '14 at 13:18
@DerekHolt: I am familiar with $\langle(1,2),(2,3),(3,4) \cdots\rangle$, not with what is written here.
Any reference for this? — johannesvalks, Jul 25 '14 at 13:50
The conjugates of $(1,2)$ under $(1,2,\ldots,n)$ are exactly $(1,2)$, $(2,3)$,$(3,4)$, etc. — Derek Holt, Jul 25 '14 at 13:52
I know it has been a long time, but I just have seen the post and I realised all the answers are wrong. The statement is NOT true for all $p$, just for $p$ prime. We know that ${(a_1 a_2), (a_1 a_2 ... a_p)}$ generate $S_p$. It is not enough proving that $\tau^k(a_1)=a_2$, you should also prove that $\tau^k$ is also a p-cycle, because if it doesn't, then is not true that $\tau^k=(a_1 a_2 ... a_p)$. The trick here is that you can only be sure $\tau^k$ is a p-cycle if $\tau$ is a p-cycle and $p$ is prime. This is not dificult to prove. — Nicolas Horvath, May 22 '20 at 08:14
Does this answer your question? Let $a$ be a p-cycle in $S_p$, and let $b$ be a transposition in $S_p$. Show $S_p$ is generated by $a$ and $b$. — AKP2002, Feb 03 '24 at 00:53

score 2 · Answer 1 · answered Jul 25 '14 at 13:36

2

As Michael stated before, you have included that information in the question itself, if $b_{i}=a_{2}$ for some $i$, then $\tau^{i-1}(a_{1})=b_{i}=a_{2}$ or $\tau^{k}(a_{1})=a_{2}$ for $k=i-1$.

answered Jul 25 '14 at 13:36

Jas

86

Show that the symmetric group $S_p=\left<\sigma, \tau\right>$, where $\sigma$ is any transposition and $\tau$ is any p-cycle and p is a prime number.

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